Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 05 Feb 2016, 19:12

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# A cylindrical tank of radius R and height H must be redesign

Author Message
TAGS:
Manager
Joined: 12 Jul 2011
Posts: 152
Concentration: Operations, Strategy
GMAT 1: 680 Q46 V37
WE: Engineering (Telecommunications)
Followers: 0

Kudos [?]: 31 [3] , given: 42

A cylindrical tank of radius R and height H must be redesign [#permalink]  24 Oct 2011, 21:17
3
KUDOS
7
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

30% (02:39) correct 70% (01:43) wrong based on 336 sessions
A cylindrical tank of radius R and height H must be redesigned to hold approximately twice as much liquid. Which of the following changes would be farthest from the new design requirements?

A. A 100% increase in R and a 50% decrease in H
B. A 30% decrease in R and a 300% increase in H
C. A 10% decrease in R and a 150% increase in H
D. A 40% increase in R and no change in H
E. A 50% increase in R and a 20% decrease in H
[Reveal] Spoiler: OA
Manager
Joined: 16 Sep 2011
Posts: 182
Concentration: Strategy, Operations
Schools: ISB '15
GMAT 1: 720 Q48 V40
GPA: 3.18
WE: Supply Chain Management (Manufacturing)
Followers: 13

Kudos [?]: 49 [1] , given: 34

Re: MGMAT Geometry #13, Redesigned Tank [#permalink]  24 Oct 2011, 22:31
1
KUDOS
ans is E
volume of cylinder=pi*r^2*H
=>reqd multiplication factor is 2
=>any change in R will cause volume to be multiplied by "square of (1+/- change in fractions)"
=>any change in H will cause vol to b multiplied by (1+/-change)
1. a 100% increase in R and a 50% decrease in H=>4*.5=2=>as required
2. a 30% decrease in R and a 300% increase in H=>.49*4=1.96
3. a 10% decrease in R and a 150% increase in H=>.81*2.5=2.025
4. a 40% increase in R and no change in H=>1.96
5. a 50% increase in R and a 20% decrease in H=>1.8=>max difference from required multiplication factor of 2

very easy qs...but my method took like 3 minutes....anyone with a faster solution?
Intern
Joined: 07 Nov 2011
Posts: 31
Followers: 0

Kudos [?]: 1 [0], given: 9

Re: MGMAT Geometry #13, Redesigned Tank [#permalink]  04 Dec 2011, 08:04
easy but time consuming...ny 1 with any easy method??
Senior Manager
Joined: 10 Nov 2010
Posts: 266
Location: India
Concentration: Strategy, Operations
GMAT 1: 520 Q42 V19
GMAT 2: 540 Q44 V21
WE: Information Technology (Computer Software)
Followers: 5

Kudos [?]: 168 [0], given: 22

Re: MGMAT Geometry #13, Redesigned Tank [#permalink]  17 Apr 2012, 22:03
vaibhav123 wrote:
easy but time consuming...ny 1 with any easy method??

Can any body explain it geometrically.
Thanks
_________________

The proof of understanding is the ability to explain it.

Math Expert
Joined: 02 Sep 2009
Posts: 31223
Followers: 5341

Kudos [?]: 62019 [2] , given: 9426

Re: MGMAT Geometry #13, Redesigned Tank [#permalink]  18 Apr 2012, 04:08
2
KUDOS
Expert's post
1
This post was
BOOKMARKED
GMATD11 wrote:
vaibhav123 wrote:
easy but time consuming...ny 1 with any easy method??

Can any body explain it geometrically.
Thanks

What do you mean geometrically?

A cylindrical tank of radius R and height H must be redesigned to hold approximately twice as much liquid. Which of the following changes would be farthest from the new design requirements?
A. A 100% increase in R and a 50% decrease in H
B. A 30% decrease in R and a 300% increase in H
C. A 10% decrease in R and a 150% increase in H
D. A 40% increase in R and no change in H
E. A 50% increase in R and a 20% decrease in H

The wording makes the question harder than it actually is.

The question basically asks: if we have a cylindrical tank of radius R and height H, then which of the following changes will give a cylindrical tank with the volume with greatest difference from the volume of $$2\pi{r^2}h$$ (twice the original).

Answer E ($$\pi{(1.5r)^2}*(0.8h)=1.8\pi{r^2}h$$) gives the greatest difference from $$2\pi{r^2}h$$.
_________________
Intern
Joined: 09 Jun 2012
Posts: 31
Followers: 0

Kudos [?]: 11 [0], given: 13

Re: A cylindrical tank of radius R and height H must be redesign [#permalink]  18 Jul 2013, 03:52
arjunbt wrote:
A cylindrical tank of radius R and height H must be redesigned to hold approximately twice as much liquid. Which of the following changes would be farthest from the new design requirements?
A. A 100% increase in R and a 50% decrease in H
B. A 30% decrease in R and a 300% increase in H
C. A 10% decrease in R and a 150% increase in H
D. A 40% increase in R and no change in H
E. A 50% increase in R and a 20% decrease in H

"Farthest from the new design requirement" denotes highest difference between "twice the volume" (- the actual requirement) and volume given by each of the answer choices (- the new design requirement). Actual required volume is 2*pi*r^2*h.
We need to calculate 2 metrics for each answer choice:
1) Volume of each of the new design requirements and
2) Difference from actual required volume (2*pi*r^2*h)
A)
Volume of new design = pi*(2r)^2 * (.50h) = 2*pi*r^2*h
Difference = 2*pi*r^2*h ~ 2*pi*r^2*h = 0
B)
Volume of new design = pi*(.7r)^2 * (4h) = 1.96 *pi*r^2*h
Difference = 1.96 *pi*r^2*h ~ 2*pi*r^2*h = 0.04 *pi*r^2*h
C)
Volume of new design = pi*(1.4r)^2 * (h) = 2.025*pi*r^2*h
Difference = 2.025*pi*r^2*h ~ 2*pi*r^2*h = 0.025 *pi*r^2*h
D)
Volume of new design = pi*(.9r)^2 * (2.5h) = 1.96*pi*r^2*h
Difference = 1.96 *pi*r^2*h ~ 2*pi*r^2*h = 0.04 *pi*r^2*h
E)
Volume of new design = pi*(1.5r)^2 * (.8h) = 1.8*pi*r^2*h
Difference = 1.8*pi*r^2*h ~ 2*pi*r^2*h = 0.2 *pi*r^2*h

Therefore, the largest difference (farthest design) is 0.2. so the correct choice is E.
Intern
Joined: 09 Jun 2012
Posts: 31
Followers: 0

Kudos [?]: 11 [0], given: 13

Re: A cylindrical tank of radius R and height H must be redesign [#permalink]  18 Jul 2013, 21:09
Also, we can follow the below technique for finding the percentage increase in volume/area.
Say if side of a square increased by 10%. The new area is (1.1*side)^2 = 1.21 (side)^2 = 1.21(Old Volume). Therefore volume increased by 21%.

Similarly, for this question, we can just multiply the numeric part before radius^2 and height after applying appropriate % increase/decrease.
The old volume is pi * r^2 *h. And the required volume is 2*pi * r^2 *h.
Working out option E -> If radius increased by 50% and height decreased by 20%, then new volume = (1.5)^2 * (.80) (Old Volume)=1.8(Old Volume). Whereas required volume is 2*(Old Volume). Hence new design is .2 farther from required design.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6216
Location: Pune, India
Followers: 1674

Kudos [?]: 9580 [3] , given: 196

Re: A cylindrical tank of radius R and height H must be redesign [#permalink]  18 Jul 2013, 22:24
3
KUDOS
Expert's post
arjunbt wrote:
A cylindrical tank of radius R and height H must be redesigned to hold approximately twice as much liquid. Which of the following changes would be farthest from the new design requirements?

A. A 100% increase in R and a 50% decrease in H
B. A 30% decrease in R and a 300% increase in H
C. A 10% decrease in R and a 150% increase in H
D. A 40% increase in R and no change in H
E. A 50% increase in R and a 20% decrease in H

The solution I am going to give has already been provided but I am still writing it down to highlight the approximations you could use to save time.

Volume must be doubled.
We know volume of a cylinder $$= \pi * r^2 * h$$
To double the volume, either you double the height 'h' OR you multiply r by $$\sqrt{2}$$ i.e. by 1.41 i.e. increase radius by a little over 40%.
OR you use a combination.

A. A 100% increase in R and a 50% decrease in H
Volume $$= \pi * 4r^2 * \frac{h}{2} = 2* \pi * r^2 * h$$ (matches the design requirement)

B. A 30% decrease in R and a 300% increase in H
Volume $$= \pi * 0.49 r^2 * 4h = \pi * \frac{r^2}{2} * 4h = 2* \pi * r^2 * h$$ (matches the design requirement)
Note that .49 is approximately 0.5 i.e. 1/2

C. A 10% decrease in R and a 150% increase in H
Volume $$= \pi * .81 r^2 * \frac{5}{2}h = 2* \pi * \frac{4}{5}r^2 * \frac{5}{2} h = 2* \pi * r^2 * h$$ (matches the design requirement)
Notice that .81 is approximately .80 i.e. 80% i.e. 4/5

D. A 40% increase in R and no change in H
We discussed it above. A 40% increase in R matches our design requirement.

E. A 50% increase in R and a 20% decrease in H
Volume $$= \pi * 2.25 r^2 * 0.8 h$$
Compare with (C). 2.5*0.8 gives you 2 so 2.25*0.8 will give you less than 2. Hence doesn't match the requirement.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 07 May 2013 Posts: 109 Followers: 0 Kudos [?]: 14 [0], given: 1 Re: A cylindrical tank of radius R and height H must be redesign [#permalink] 01 Dec 2013, 04:35 Karishma, I took R=10 and H=10 and proceeded but I am unable to arrive at a constructive answer. Can you please explain how to proceed using the above values and using substitution method. Thanks in advance . Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6216 Location: Pune, India Followers: 1674 Kudos [?]: 9580 [0], given: 196 Re: A cylindrical tank of radius R and height H must be redesign [#permalink] 01 Dec 2013, 20:22 Expert's post madn800 wrote: Karishma, I took R=10 and H=10 and proceeded but I am unable to arrive at a constructive answer. Can you please explain how to proceed using the above values and using substitution method. Thanks in advance . R = 10 H = 10 We know volume of a cylinder$$= \pi * r^2 * h = 1000\pi$$ A. A 100% increase in R and a 50% decrease in H (R = 20, H = 5) Volume =$$\pi * 400 * 5 = 2000\pi$$(matches the design requirement) B. A 30% decrease in R and a 300% increase in H (R = 7, H = 40) Volume =$$\pi * 49 * 40 = 1960\pi$$ (approximately matches the design requirement) Note that 49 is approximately 50 C. A 10% decrease in R and a 150% increase in H (R = 9, H = 25) Volume$$= \pi * 81 * 25 = 2025\pi$$ (approximately matches the design requirement) Notice that 81 is approximately 80 D. A 40% increase in R and no change in H (R = 14) Volume$$= \pi * 196 * 10 = 1960 \pi$$ (approximately matches the design requirement) E. A 50% increase in R and a 20% decrease in H (R = 15, H = 8) Volume $$= \pi * 225 * 8 = 1800\pi$$ (Does not match) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Manager
Joined: 17 Mar 2014
Posts: 92
Followers: 0

Kudos [?]: 33 [1] , given: 194

A cylindrical tank of radius R and height H must be redesign [#permalink]  03 Nov 2014, 10:20
1
KUDOS
I liked below solution. Just thought of posting here, it may help others. Credit goes to http://www.beatthegmat.com/new-design-r ... tml#231919
It is almost similar to Karishma and Bunuel's solution.

The new design requirements need the final volume to be 2 π R^2 H from π R^2 H, or it needs to be doubled (2 times).

QUICK CHECK is here

A. 2^2 X 0.5 = 2, (2 - 2 = 0)
B. 0.7^2 X 4 = 1.96, (2 - 1.96 = 0.04)
C. 0.9^2 X 2.5 = 2.025, (2.025 - 2 = 0.025)
D. 1.4^2 X 1 = 1.96, (2 - 1.96 = 0.04)
E. 1.5^2 X 0.8 = 1.8, (2 - 1.8 = 0.2), this is the GREATEST DEVIATION

Regards,
Ammu
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 8149
Followers: 417

Kudos [?]: 110 [0], given: 0

Re: A cylindrical tank of radius R and height H must be redesign [#permalink]  04 Dec 2015, 07:12
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Intern
Joined: 07 Oct 2015
Posts: 9
Followers: 0

Kudos [?]: 0 [0], given: 2

Re: A cylindrical tank of radius R and height H must be redesign [#permalink]  04 Dec 2015, 09:00
Can't you just use the formula, and multiply it pretty quickly? Took me less than 2 minutes if you just use the percentage increases.

A) 100% increase to R = 2R and 50% decrease to H = .5 H so formula is pi*r^2*h
so 2^2*.5=2

Repeat with each solution and you get E at 1.8.

This worked for me, but is this correct?
Re: A cylindrical tank of radius R and height H must be redesign   [#permalink] 04 Dec 2015, 09:00
Similar topics Replies Last post
Similar
Topics:
4 A cylindrical water tank has a diameter of 14 meters and a height of 8 02 Jul 2015, 00:35
1 A cylindrical tank, with radius and height both of 10 feet, is to be r 1 29 Oct 2014, 07:18
The radius of a cylindrical water tank is reduced by 50%. Ho 2 19 Dec 2013, 11:05
4 A cylindrical can has a radius of 4 centimeters and a height 3 26 Sep 2013, 02:47
6 A cylindrical tank with radius 3 meters is filled with a solution. The 18 31 Jan 2011, 21:30
Display posts from previous: Sort by